Answer
The lower angular frequency is $~~219~rad/s$
Work Step by Step
The current amplitude at resonance is $I = \frac{\mathscr{E}_m}{R}$
In general, the current amplitude is $I = \frac{\mathscr{E}_m}{Z}$
If the current amplitude is half the maximum value, then $Z = 2R$
We can find $\omega_d$:
$Z = \sqrt{R^2+(X_L-X_C)^2}$
$2R = \sqrt{R^2+(X_L-X_C)^2}$
$4R^2 = R^2+(X_L-X_C)^2$
$3R^2 = (X_L-X_C)^2$
$\sqrt{3}~R = X_L-X_C$
$\sqrt{3}~R = \omega_d~L-\frac{1}{\omega_d~C}$
$\sqrt{3}~R~\omega_d = \omega_d^2~L-\frac{1}{C}$
$\omega_d^2~L-\sqrt{3}~R~\omega_d-\frac{1}{C} = 0$
$(1.00~H)~\omega_d^2-\sqrt{3}~(5.00~\Omega)~\omega_d-\frac{1}{20.0\times 10^{-6}~F} = 0$
$(1.00~H)~\omega_d^2-(8.66~\Omega)~\omega_d-50,000~F^{-1} = 0$
We can use the quadratic formula:
$\omega_d = \frac{8.66\pm \sqrt{(-8.66)^2-(4)(1.00)(-50,000)}}{(2)(1.00)}$
$\omega_d = \frac{8.66\pm 447.3}{2.00}$
$\omega_d = -219~rad/s, 228~rad/s$
Since the circuit is oscillating, we can say that the lower angular frequency is $~~219~rad/s$
